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Proof step 11 from proof sgn-lastn-01 (previous step)

bits0 ≠ [] ∧ (bits0.first ↔ boolvar), n ≤ # sgn(bits0), ∀ n0. ¬ # sgn(bits0) < n0 → bits0 = [] ∨ lastn(n0, sgn(bits0)) = lastn(n0, bits0) ⊦ lastn(n, sgn(bits0)) = lastn(n, boolvar ' + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n ≤ # sgn(bits0), ∀ n0. ¬ # sgn(bits0) < n0 → lastn(n0, sgn(bits0)) = lastn(n0, bits0) ⊦ lastn(n, sgn(bits0)) = lastn(n, bits0.first ' + bits0), bits0 = []

Proof step 11 from proof sgn-lastn-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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