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Proof step 6 from proof inj (previous step)

n→i(m) = n→i(n0) + 1, n→i(m - 1) = n→i(n0) → m - 1 = n0, ∀ m. n→i(m) = n→i(n0) → m = n0 ⊦ m = n0 + 1

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0]

Additional information

Used simplifier rules: nat2int-07: ⊦ 0 ≠ n→i(m) + 1;%succ-01: ⊦ n→i(m + 1) = n→i(m) + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ m + 1 ≠ 0; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0