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Proof step 3 from proof sgn-lastn-01 (previous step)

∀ n. n ≤ # sgn(bits0) → bits0 = [] ∨ lastn(n, sgn(bits0)) = lastn(n, bits0), n ≤ # sgn(boolvar + bits0) ⊦ boolvar + bits0 = [], lastn(n, sgn(boolvar + bits0)) = lastn(n, boolvar + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;