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Proof step 6 from proof sgn-one-01 (previous step)

sgn(boolvar ' + bits0) = true ', sgn(bits0) = true ' ∧ false ∈ bits0 → bits0 = [] ⊦ ¬ false ∈ bits0 ∧ boolvar

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:
¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), sgn(boolvar ' + bits0) = true ', sgn(bits0) = true ' ∧ false ∈ bits0 → bits0 = [] ⊦ ¬ false ∈ bits0 ∧ boolvar
Proof step 15:
bits0 ≠ [] ∧ (bits0.first ↔ boolvar), sgn(bits0) = true ', sgn(bits0) = true ' ∧ false ∈ bits0 → bits0 = [] ⊦ ¬ false ∈ bits0 ∧ boolvar

Rule argument

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

Additional information

Used terms: bits0, boolvar