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Proof step 71 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits2 + boolvar '), bits3 + boolvar0 ')) = fillsgn(n, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'), bits2 = [], bits3 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 72:
⊦ n ≠ 0 ∧ bits ≠ [] → fillsgn(n, bits + boolvar ') = fillsgn(n -1, bits) + boolvar
Proof step 73:
¬(n ≠ 0 ∧ bits2 ≠ []), ∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits2 + boolvar '), bits3 + boolvar0 ')) = fillsgn(n, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'), bits2 = [], bits3 = []
Proof step 76:
n ≠ 0 ∧ bits2 ≠ [], ∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n -1, bits2) + boolvar, bits3 + boolvar0 ')) = fillsgn(n, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'), bits2 = [], bits3 = []

Rule argument

seq: ⊦ n ≠ 0 ∧ bits ≠ [] → fillsgn(n, bits + boolvar ') = fillsgn(n -1, bits) + boolvar subst: [boolvar, bits, n] → [boolvar, bits2, n] rotate: F spec: bitopsaux inst: name: fill-last

Additional information

Used terms: boolvar, bits2, n