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Proof step 8 from proof sgn-lastn-01 (previous step)

bits0 = boolvar0 + bits, n ≤ # sgn(boolvar ' + bits0), ∀ n0. ¬ # sgn(bits0) < n0 → lastn(n0, sgn(bits0)) = lastn(n0, bits0) ⊦ lastn(n, sgn(boolvar ' + bits0)) = lastn(n, boolvar ' + bits0), boolvar0 ↔ boolvar, bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n ≤ # sgn(boolvar ' + boolvar0 ' + bits), ∀ n0. ¬ # sgn(boolvar0 ' + bits) < n0 → lastn(n0, sgn(boolvar0 ' + bits)) = lastn(n0, boolvar0 ' + bits) ⊦ boolvar0 ↔ boolvar, lastn(n, sgn(boolvar ' + boolvar0 ' + bits)) = lastn(n, boolvar ' + boolvar0 ' + bits)

Rule argument

Additional information

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