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Proof step 21 from proof sgn-ex-01 (previous step)

sgn(mklist(false, n) + true ' + bits) = false ' + true ' + bits ⊦ n = 0, mklist(false, n) = false ' + mklist(false, n - 1), ∃ boolvar2, n0, boolvar3, bits1. true ' + mklist(false, n) + true ' + bits = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: sgn-mklist-02: ⊦ sgn(false ' + mklist(false, n) + bits) = sgn(false ' + bits);%sgn-base-02: ⊦ sgn(false ' + true ' + bits) = false ' + true ' + bits;% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);%n ≠ 0 ⊦ mklist(false, (* n)) = false ' + mklist(false, n -1);% ⊦ mklist(boolvar, 0) = [];% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ m + 1 ≠ 0;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m