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Proof step 5 from proof enc-zero-01 (previous step)

2 * i→n(i) - 1 = 0, i < 0 ⊦ i = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [2 * i→n(i), m] flags:

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%one-01: ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%two-zero-01: ⊦ 2 * m = 0 ↔ m = 0; from specification nat-mult%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0; from specification intnat%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: 2 * i→n(i), m