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Proof step 3 from proof prefixn-add (previous step)

⊦ prefixn(n, a ' + x0) = prefixn(n, a ' + x1) ↔ prefixn(n - 1, x0) = prefixn(n - 1, x1), n = 0

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] → [n, m]

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ prefixn(n + 1, a ' + x) = a ' + prefixn(n, x); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x = a ' + y ↔ x = y; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m