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Proof step 21 from proof and-insgn (previous step)

∀ bits1, bits, boolvar. # bits1 < #(boolvar0 + bits2) → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits = true ', bits = false ', boolvar ∈ bitsand(sgn(bits), boolvar0 ' + bits2) ↔ boolvar ∈ bitsand(bits, boolvar0 ' + bits2), bits = [], boolvar0 ' + bits2 = true ', boolvar0 ' + bits2 = false ', bits2 ≠ [] ∧ bits ≠ bits.first '

User applied rule insert elim lemma

And closed the goal!

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits, boolvar1, bits0] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ bits + [] = bits;% ⊦ sgn(boolvar ') = boolvar ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar1, bits0