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Proof step 5 from proof fill-one-02 (previous step)

0 < i ⊦ boolvar ' + mklist(boolvar, i→n(i) - 1) = mklist(boolvar, i→n(i))

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] → [i→n(i), m] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ mklist(boolvar, 1) = boolvar ';% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: i→n(i), m