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Proof step 6 from proof sgn-ex (previous step)

∀ boolvar. ∃ n. boolvar ' + bits = mklist(boolvar, n) + sgn(boolvar ' + bits) ⊦ ∃ n. boolvar0 ' + boolvar ' + bits = mklist(boolvar0, n) + sgn(boolvar0 ' + boolvar ' + bits)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
bits ≠ [], bits.first ↔ boolvar ⊦ sgn(boolvar ' + bits) = sgn(bits)
Proof step 8:
¬(boolvar ' + bits ≠ [] ∧ ((boolvar ' + bits).first ↔ boolvar0)), ∀ boolvar. ∃ n. boolvar ' + bits = mklist(boolvar, n) + sgn(boolvar ' + bits) ⊦ ∃ n. boolvar0 ' + boolvar ' + bits = mklist(boolvar0, n) + sgn(boolvar0 ' + boolvar ' + bits)
Proof step 10:
boolvar ' + bits ≠ [] ∧ ((boolvar ' + bits).first ↔ boolvar0), ∀ boolvar. ∃ n. boolvar ' + bits = mklist(boolvar, n) + sgn(boolvar ' + bits) ⊦ ∃ n. boolvar0 ' + boolvar ' + bits = mklist(boolvar0, n) + sgn(boolvar ' + bits)

Rule argument

name: sgn-rec-01 rotate: F subst: [bits, boolvar] → [boolvar ' + bits, boolvar0]

Additional information

Used terms: boolvar ' + bits, boolvar0