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Proof step 4 from proof sgn-lastn-01 (previous step)

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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
bits ≠ [], bits.first ↔ boolvar ⊦ sgn(boolvar ' + bits) = sgn(bits)
Proof step 6:
¬(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)
Proof step 11:
bits0 ≠ [] ∧ (bits0.first ↔ boolvar), n ≤ # sgn(bits0), ∀ n0. ¬ # sgn(bits0) < n0 → bits0 = [] ∨ lastn(n0, sgn(bits0)) = lastn(n0, bits0) ⊦ lastn(n, sgn(bits0)) = lastn(n, boolvar ' + bits0)

Rule argument

name: sgn-rec-01 rotate: F subst: [bits, boolvar] → [bits0, boolvar]

Additional information

Used terms: bits0, boolvar