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Proof step 96 from proof bxor-flip-one-bit (previous step)

int2bits(i) = false ' + (false ' + bits) + boolvar ', false ∈ int2bits(i), 0 < i, 2 < i ⊦ i = 0, (sgn(false ' + bits + (¬ boolvar)')[2 + # bits] ↔ ¬ boolvar) ∧ sgn(false ' + bits + (¬ boolvar)')[2 + # bits, boolvar] = int2bits(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 97:
⊦ (bits.first ↔ bits.rest.first) ∧ 1 < # bits → int2bits(i) ≠ bits

Rule argument

seq: ⊦ (bits.first ↔ bits.rest.first) ∧ 1 < # bits → int2bits(i) ≠ bits subst: [bits, i] → [false ' + (false ' + bits) + boolvar ', i] rotate: F spec: int2bits inst: name: int2bits-leading-sgn

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + n = n + m;% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0); Used terms: false ' + (false ' + bits) + boolvar ', i