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Proof step 6 from proof sgn-lastn-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # sgn(boolvar ' + bits0), ∀ n0. ¬ # sgn(bits0) < n0 → lastn(n0, sgn(bits0)) = lastn(n0, bits0) ⊦ lastn(n, sgn(boolvar ' + bits0)) = lastn(n, boolvar ' + bits0), bits0.first ↔ boolvar, bits0 = []

Rule argument

Additional information

Used simplifier rules: sgn-one: ⊦ sgn(boolvar ') = boolvar ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;