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Proof step 12 from proof fill-last (previous step)

∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = fillsgn(m0 + 1, bits) + boolvar '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits)
Proof step 14:
¬(bits ≠ [] ∧ # bits < m0 + 1), ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = fillsgn(m0 + 1, bits) + boolvar '
Proof step 20:
bits ≠ [] ∧ # bits < m0 + 1, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = (bits.first ' + fillsgn((m0 + 1)-1, bits)) + boolvar '

Rule argument

name: fill-rec rotate: F subst: [bits, n] → [bits, m0 + 1]

Additional information

Used terms: bits, m0 + 1