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Course contents

  • How to Use This Course
  • Reading Paths
  • Glossary
  • Metrics Reference
  • Formal Cheat Sheet
  • Formal Foundations19
    • Formal versus Simulation
    • Transition Systems, Invariants and the Three Property Roles
    • Reset, Initial State and the First Cycle
    • Cone of Influence
    • The Formal Tool Flow, Vendor-Neutral
      • Anatomy of a Formal Harness
      • Tasks, Engines and Time Limits
      • Reading a Result Report
      • Model Fidelity and Unsupported Constructs
  • Properties and SVA16
  • Property Development and Constraints19
  • Proof Engines and Convergence23
  • Formal Applications20
  • Equivalence Checking14
  • Formal Coverage and Signoff15
  • Methodology and Debug13
  • Interview Prep15
  • Capstone: A Chip-Block Verification Lab24
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←The Formal Tool Flow, Vendor-Neutral
  1. Topics
  2. Formal Verification
  3. Formal Foundations
  4. The Formal Tool Flow, Vendor-Neutral
  5. Tasks, Engines and Time Limits
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    Part 1 of 10, lesson 17 of 19Part 1 of 10/Formal Foundations/Lesson 17 of 19All levels

    Tasks, Engines and Time Limits

    What a task, an engine, a depth and a time limit each control, shown on one design at two depths.

    Why it matters

    An engineer doubled the time limit on a stalled run and waited twice as long for the same inconclusive answer. The run did not need more time; it needed a different engine, a deeper bound, or a split property set. Depth, engine, time and splitting are four different controls, and turning the wrong one wastes the run. This lesson shows what each does on one design.

    Job of this lesson: define task, engine, depth and time limit in any formal tool, and show what changing each one does.

    The core idea

    Control

    What it is

    What changing it does

    Task

    The question: prove unboundedly, check boundedly to N, or search for a witness

    Changes what a result means: a bounded pass is not a proof.

    Engine

    The method that answers it: unrolling for bounded checks, the induction step for proofs

    A different engine answers a different question.

    Depth

    How far the unrolling reaches

    Looks further; costs clauses linearly.

    Time limit

    How long the solver may work

    Helps only a run that was making progress.

    Splitting

    One property per run

    Narrows each cone.

    Worked example: one miter at two depths

    The running design is a three-stage arithmetic pipeline beside its retimed twin, compared output by output in a miter (a circuit that feeds both the same inputs and raises a flag when the outputs differ) with 48 design state bits. The bug variant of the twin ignores bit 3 of one operand, and the outputs diverge at step 4, when the first input vector reaches the output. Depth 4 passes and depth 5 fails on the same design; the good twin proves.

    pipe3.sv
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    // formal: bmc 4 expect pass slang top pipe3_eq define BUG=1
    `define FORMAL
    // pipe3: three-stage arithmetic pipeline, y = ((a + b)[3:0] * c[3:0]) + 1 (8 bits),
    // registers: stage 1 {s = a + b, c1 = c[3:0]}, stage 2 {p = s[3:0] * c1}, stage 3 {y = p + 1}.
    // Latency 3 cycles (y at cycle t+3 uses a, b, c of cycle t). Synchronous active-high
    // reset clears every register. Only a[3:0], b[3:0], c[3:0] influence y.
    // pipe3_retimed: same function and latency, registers moved across the multiplier:
    // stage 1 {s, c1} (same), stage 2 {s2 = s[3:0], c2 = c1} (the multiplier moved to
    // stage 3), stage 3 {y = s2 * c2 + 1}. BUG=1 injects a bug in the retimed twin: the
    // multiplier ignores bit 3 of c2.
    // pipe3_eq: miter of pipe3 against pipe3_retimed with common inputs; BUG is passed to the twin.
    module pipe3 #(parameter CHECK = 1) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output reg  [7:0] y,
      output wire [7:0] dbg_p          // stage-2 register, exposed for the miter's helper invariant
    );
      reg [7:0] s;
      reg [3:0] c1;
      reg [7:0] p;
      assign dbg_p = p;
      always @(posedge clk) begin
        if (rst) begin s <= 8'd0; c1 <= 4'd0; p <= 8'd0; y <= 8'd0; end
        else begin
          s  <= a + b;
          c1 <= c[3:0];
          p  <= s[3:0] * c1;
          y  <= p + 8'd1;
        end
      end
    
    `ifdef FORMAL
      // CHECK=0 switches the properties off (used when pipe3 sits inside the miter)
      generate if (CHECK) begin : g_chk
      reg init = 1'b1;
      always @(posedge clk) init <= 1'b0;
      always @(*) if (init) assume (rst);
    
      // hist counts clean cycles since reset (saturates at 3): y is only specified after 3
      reg [1:0] hist;
      always @(posedge clk) begin
        if (rst) hist <= 2'd0;
        else if (hist != 2'd3) hist <= hist + 2'd1;
      end
    
      // input history, as plain registers (works in every front end)
      reg [3:0] sa1, sb1, sc1, sa2, sb2, sc2, sa3, sb3, sc3;
      always @(posedge clk) begin
        sa1 <= a[3:0]; sb1 <= b[3:0]; sc1 <= c[3:0];
        sa2 <= sa1;    sb2 <= sb1;    sc2 <= sc1;
        sa3 <= sa2;    sb3 <= sb2;    sc3 <= sc2;
      end
      wire [3:0] sum3 = sa3 + sb3;       // low 4 bits of a + b three cycles ago
    
      // functional property with latency 3, checked only after 3 clean cycles
      always @(*) if (!init && hist == 2'd3) a_latency3: assert (y == (sum3 * sc3) + 8'd1);
      // after reset the output register is 0
      always @(posedge clk) if (!init && $past(rst)) a_reset_zero: assert (y == 8'd0);
      always @(*) if (!init) c_y_nonzero: cover (hist == 2'd3 && y == 8'd7);
      end endgenerate
    `endif
    endmodule
    
    module pipe3_retimed #(parameter BUG = 0) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output reg  [7:0] y,
      output wire [3:0] dbg_s2,        // stage-2 registers, exposed for the miter's helper invariant
      output wire [3:0] dbg_c2
    );
      reg [7:0] s;
      reg [3:0] c1;
      reg [3:0] s2, c2;
      assign dbg_s2 = s2;
      assign dbg_c2 = c2;
      always @(posedge clk) begin
        if (rst) begin s <= 8'd0; c1 <= 4'd0; s2 <= 4'd0; c2 <= 4'd0; y <= 8'd0; end
        else begin
          s  <= a + b;
          c1 <= c[3:0];
          s2 <= s[3:0];
          c2 <= c1;
          y  <= s2 * ((BUG == 1) ? {1'b0, c2[2:0]} : c2) + 8'd1;
        end
      end
    endmodule
    
    module pipe3_eq #(parameter BUG = 0, parameter HELPER = 1) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output wire       same
    );
      wire [7:0] y1, y2, p_ref;
      wire [3:0] s2_ret, c2_ret;
      pipe3 #(.CHECK(0))     u_ref (.clk(clk), .rst(rst), .a(a), .b(b), .c(c), .y(y1), .dbg_p(p_ref));
      pipe3_retimed #(.BUG(BUG)) u_ret (.clk(clk), .rst(rst), .a(a), .b(b), .c(c), .y(y2), .dbg_s2(s2_ret), .dbg_c2(c2_ret));
      assign same = (y1 == y2);
    
    `ifdef FORMAL
      reg init = 1'b1;
      always @(posedge clk) init <= 1'b0;
      always @(*) if (init) assume (rst);
      always @(*) if (!init) a_equiv: assert (same);
      // helper invariant across the moved cut: the reference's stage-2 register p equals the
      // twin's s2 * c2 (the multiplier result computed one stage later). Not needed if k >= 2.
      generate if (HELPER) begin : g_helper
        always @(*) if (!init) h_cut: assert (p_ref == s2_ret * c2_ret);
      end endgenerate
      always @(*) if (!init) c_nonzero: cover (y1 == 8'd7 && y2 == 8'd7);
    `endif
    endmodule
    pipe3.sv
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    // formal: bmc 5 expect fail slang top pipe3_eq define BUG=1
    `define FORMAL
    // pipe3: three-stage arithmetic pipeline, y = ((a + b)[3:0] * c[3:0]) + 1 (8 bits),
    // registers: stage 1 {s = a + b, c1 = c[3:0]}, stage 2 {p = s[3:0] * c1}, stage 3 {y = p + 1}.
    // Latency 3 cycles (y at cycle t+3 uses a, b, c of cycle t). Synchronous active-high
    // reset clears every register. Only a[3:0], b[3:0], c[3:0] influence y.
    // pipe3_retimed: same function and latency, registers moved across the multiplier:
    // stage 1 {s, c1} (same), stage 2 {s2 = s[3:0], c2 = c1} (the multiplier moved to
    // stage 3), stage 3 {y = s2 * c2 + 1}. BUG=1 injects a bug in the retimed twin: the
    // multiplier ignores bit 3 of c2.
    // pipe3_eq: miter of pipe3 against pipe3_retimed with common inputs; BUG is passed to the twin.
    module pipe3 #(parameter CHECK = 1) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output reg  [7:0] y,
      output wire [7:0] dbg_p          // stage-2 register, exposed for the miter's helper invariant
    );
      reg [7:0] s;
      reg [3:0] c1;
      reg [7:0] p;
      assign dbg_p = p;
      always @(posedge clk) begin
        if (rst) begin s <= 8'd0; c1 <= 4'd0; p <= 8'd0; y <= 8'd0; end
        else begin
          s  <= a + b;
          c1 <= c[3:0];
          p  <= s[3:0] * c1;
          y  <= p + 8'd1;
        end
      end
    
    `ifdef FORMAL
      // CHECK=0 switches the properties off (used when pipe3 sits inside the miter)
      generate if (CHECK) begin : g_chk
      reg init = 1'b1;
      always @(posedge clk) init <= 1'b0;
      always @(*) if (init) assume (rst);
    
      // hist counts clean cycles since reset (saturates at 3): y is only specified after 3
      reg [1:0] hist;
      always @(posedge clk) begin
        if (rst) hist <= 2'd0;
        else if (hist != 2'd3) hist <= hist + 2'd1;
      end
    
      // input history, as plain registers (works in every front end)
      reg [3:0] sa1, sb1, sc1, sa2, sb2, sc2, sa3, sb3, sc3;
      always @(posedge clk) begin
        sa1 <= a[3:0]; sb1 <= b[3:0]; sc1 <= c[3:0];
        sa2 <= sa1;    sb2 <= sb1;    sc2 <= sc1;
        sa3 <= sa2;    sb3 <= sb2;    sc3 <= sc2;
      end
      wire [3:0] sum3 = sa3 + sb3;       // low 4 bits of a + b three cycles ago
    
      // functional property with latency 3, checked only after 3 clean cycles
      always @(*) if (!init && hist == 2'd3) a_latency3: assert (y == (sum3 * sc3) + 8'd1);
      // after reset the output register is 0
      always @(posedge clk) if (!init && $past(rst)) a_reset_zero: assert (y == 8'd0);
      always @(*) if (!init) c_y_nonzero: cover (hist == 2'd3 && y == 8'd7);
      end endgenerate
    `endif
    endmodule
    
    module pipe3_retimed #(parameter BUG = 0) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output reg  [7:0] y,
      output wire [3:0] dbg_s2,        // stage-2 registers, exposed for the miter's helper invariant
      output wire [3:0] dbg_c2
    );
      reg [7:0] s;
      reg [3:0] c1;
      reg [3:0] s2, c2;
      assign dbg_s2 = s2;
      assign dbg_c2 = c2;
      always @(posedge clk) begin
        if (rst) begin s <= 8'd0; c1 <= 4'd0; s2 <= 4'd0; c2 <= 4'd0; y <= 8'd0; end
        else begin
          s  <= a + b;
          c1 <= c[3:0];
          s2 <= s[3:0];
          c2 <= c1;
          y  <= s2 * ((BUG == 1) ? {1'b0, c2[2:0]} : c2) + 8'd1;
        end
      end
    endmodule
    
    module pipe3_eq #(parameter BUG = 0, parameter HELPER = 1) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output wire       same
    );
      wire [7:0] y1, y2, p_ref;
      wire [3:0] s2_ret, c2_ret;
      pipe3 #(.CHECK(0))     u_ref (.clk(clk), .rst(rst), .a(a), .b(b), .c(c), .y(y1), .dbg_p(p_ref));
      pipe3_retimed #(.BUG(BUG)) u_ret (.clk(clk), .rst(rst), .a(a), .b(b), .c(c), .y(y2), .dbg_s2(s2_ret), .dbg_c2(c2_ret));
      assign same = (y1 == y2);
    
    `ifdef FORMAL
      reg init = 1'b1;
      always @(posedge clk) init <= 1'b0;
      always @(*) if (init) assume (rst);
      always @(*) if (!init) a_equiv: assert (same);
      // helper invariant across the moved cut: the reference's stage-2 register p equals the
      // twin's s2 * c2 (the multiplier result computed one stage later). Not needed if k >= 2.
      generate if (HELPER) begin : g_helper
        always @(*) if (!init) h_cut: assert (p_ref == s2_ret * c2_ret);
      end endgenerate
      always @(*) if (!init) c_nonzero: cover (y1 == 8'd7 && y2 == 8'd7);
    `endif
    endmodule
    pipe3.sv
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    // formal: prove 5 expect pass slang top pipe3_eq
    `define FORMAL
    // pipe3: three-stage arithmetic pipeline, y = ((a + b)[3:0] * c[3:0]) + 1 (8 bits),
    // registers: stage 1 {s = a + b, c1 = c[3:0]}, stage 2 {p = s[3:0] * c1}, stage 3 {y = p + 1}.
    // Latency 3 cycles (y at cycle t+3 uses a, b, c of cycle t). Synchronous active-high
    // reset clears every register. Only a[3:0], b[3:0], c[3:0] influence y.
    // pipe3_retimed: same function and latency, registers moved across the multiplier:
    // stage 1 {s, c1} (same), stage 2 {s2 = s[3:0], c2 = c1} (the multiplier moved to
    // stage 3), stage 3 {y = s2 * c2 + 1}. BUG=1 injects a bug in the retimed twin: the
    // multiplier ignores bit 3 of c2.
    // pipe3_eq: miter of pipe3 against pipe3_retimed with common inputs; BUG is passed to the twin.
    module pipe3 #(parameter CHECK = 1) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output reg  [7:0] y,
      output wire [7:0] dbg_p          // stage-2 register, exposed for the miter's helper invariant
    );
      reg [7:0] s;
      reg [3:0] c1;
      reg [7:0] p;
      assign dbg_p = p;
      always @(posedge clk) begin
        if (rst) begin s <= 8'd0; c1 <= 4'd0; p <= 8'd0; y <= 8'd0; end
        else begin
          s  <= a + b;
          c1 <= c[3:0];
          p  <= s[3:0] * c1;
          y  <= p + 8'd1;
        end
      end
    
    `ifdef FORMAL
      // CHECK=0 switches the properties off (used when pipe3 sits inside the miter)
      generate if (CHECK) begin : g_chk
      reg init = 1'b1;
      always @(posedge clk) init <= 1'b0;
      always @(*) if (init) assume (rst);
    
      // hist counts clean cycles since reset (saturates at 3): y is only specified after 3
      reg [1:0] hist;
      always @(posedge clk) begin
        if (rst) hist <= 2'd0;
        else if (hist != 2'd3) hist <= hist + 2'd1;
      end
    
      // input history, as plain registers (works in every front end)
      reg [3:0] sa1, sb1, sc1, sa2, sb2, sc2, sa3, sb3, sc3;
      always @(posedge clk) begin
        sa1 <= a[3:0]; sb1 <= b[3:0]; sc1 <= c[3:0];
        sa2 <= sa1;    sb2 <= sb1;    sc2 <= sc1;
        sa3 <= sa2;    sb3 <= sb2;    sc3 <= sc2;
      end
      wire [3:0] sum3 = sa3 + sb3;       // low 4 bits of a + b three cycles ago
    
      // functional property with latency 3, checked only after 3 clean cycles
      always @(*) if (!init && hist == 2'd3) a_latency3: assert (y == (sum3 * sc3) + 8'd1);
      // after reset the output register is 0
      always @(posedge clk) if (!init && $past(rst)) a_reset_zero: assert (y == 8'd0);
      always @(*) if (!init) c_y_nonzero: cover (hist == 2'd3 && y == 8'd7);
      end endgenerate
    `endif
    endmodule
    
    module pipe3_retimed #(parameter BUG = 0) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output reg  [7:0] y,
      output wire [3:0] dbg_s2,        // stage-2 registers, exposed for the miter's helper invariant
      output wire [3:0] dbg_c2
    );
      reg [7:0] s;
      reg [3:0] c1;
      reg [3:0] s2, c2;
      assign dbg_s2 = s2;
      assign dbg_c2 = c2;
      always @(posedge clk) begin
        if (rst) begin s <= 8'd0; c1 <= 4'd0; s2 <= 4'd0; c2 <= 4'd0; y <= 8'd0; end
        else begin
          s  <= a + b;
          c1 <= c[3:0];
          s2 <= s[3:0];
          c2 <= c1;
          y  <= s2 * ((BUG == 1) ? {1'b0, c2[2:0]} : c2) + 8'd1;
        end
      end
    endmodule
    
    module pipe3_eq #(parameter BUG = 0, parameter HELPER = 1) (
      input  wire       clk,
      input  wire       rst,
      input  wire [7:0] a,
      input  wire [7:0] b,
      input  wire [7:0] c,
      output wire       same
    );
      wire [7:0] y1, y2, p_ref;
      wire [3:0] s2_ret, c2_ret;
      pipe3 #(.CHECK(0))     u_ref (.clk(clk), .rst(rst), .a(a), .b(b), .c(c), .y(y1), .dbg_p(p_ref));
      pipe3_retimed #(.BUG(BUG)) u_ret (.clk(clk), .rst(rst), .a(a), .b(b), .c(c), .y(y2), .dbg_s2(s2_ret), .dbg_c2(c2_ret));
      assign same = (y1 == y2);
    
    `ifdef FORMAL
      reg init = 1'b1;
      always @(posedge clk) init <= 1'b0;
      always @(*) if (init) assume (rst);
      always @(*) if (!init) a_equiv: assert (same);
      // helper invariant across the moved cut: the reference's stage-2 register p equals the
      // twin's s2 * c2 (the multiplier result computed one stage later). Not needed if k >= 2.
      generate if (HELPER) begin : g_helper
        always @(*) if (!init) h_cut: assert (p_ref == s2_ret * c2_ret);
      end endgenerate
      always @(*) if (!init) c_nonzero: cover (y1 == 8'd7 && y2 == 8'd7);
    `endif
    endmodule

    Depth

    Sees step 4?

    Verdict on the bug variant

    Reading

    4

    No

    Pass

    Bounded to 4: the divergent outputs are one step outside.

    5

    Yes

    Counterexample at step 4

    The first input vector reaches the output.

    The script prices the two depths for illustrative sizes (2,000 gates and the miter's 48 flops, labelled illustrative): unrolled size is gates times three times depth plus one equality per flop per step. Depth 5 costs a quarter more than depth 4, and ten times the depth costs ten times the clauses. The check runs on the formula and on three mutants, all caught.

    example.py
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    def unrolled(gates, flops, depth):
        return gates * 3 * depth + flops * depth
    
    def check(fn):
        assert fn(2000, 48, 4) == 24000 + 192
        assert fn(2000, 48, 5) == 30000 + 240
        assert fn(2000, 48, 40) == 10 * fn(2000, 48, 4)
        assert fn(2000, 0, 5) == 30000
        assert fn(0, 48, 5) == 240
    
    check(unrolled)
    mutants = [
        lambda g, f, d: g * 3 * d,
        lambda g, f, d: g * d + f * d,
        lambda g, f, d: g * 3 + f * d,
    ]
    for mutant in mutants:
        caught = False
        try:
            check(mutant)
        except AssertionError:
            caught = True
        assert caught
    print("PASS: depth 4 costs 24,192 clauses and depth 5 costs 30,240 in the illustrative miter; all three mutants are caught")

    What the review looks for

    • Each run names its task, engine, depth and time limit beside the verdict; a bounded pass states its depth.

    • Depth changes are justified by trigger length plus latency, not by round numbers; slow runs are split by property before the time limit is raised.

    Verify and debug

    Read the status table per property at two depths before touching any control: if the deeper bound flips the verdict, the question was depth, not effort. Raise the time limit only when the solver reports progress; otherwise reduce, split or strengthen. Never tighten assumptions to buy speed without a written justification.

    More time instead of more thought. Symptom: repeated time-outs with no depth progress. Cause: treating the limit as a fix rather than a budget. Fix: diagnose from cone size, depth reached and engine progress, then reduce, split or strengthen. Found by the stalled depth across longer runs.

    “Your run timed out. What do you change?” Answer in order: diagnose, then reduce, split, strengthen; more depth or a different engine only with a reason, more time only with progress, tighter assumptions last with a written justification. Quote the miter pair: depth 4 passes and depth 5 fails, so the control that mattered was depth.

    Check yourself Depth 4 passes and depth 5 fails on the same design. Which control was wrong?

    Depth: the trigger needs step 4, so depth 4 cannot see it. Raising the time limit at depth 4 would change nothing.

    Check yourself When is raising the time limit the right move?

    When the solver reports steady progress and the cone and depth are already right: the run needs budget, not a new question.

    Related topics

    • Next: Reading a Result Report

    • Later: Bounded versus Unbounded Proofs

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    • Model Fidelity and Unsupported Constructs