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Proof step 13 from proof put-at (previous step)

m = m0 + 1, m ≤ # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ (a0 ' + x0[m0, a])[n] = (a0 ' + x0)[n], m = n, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
m0 < # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ m0 + 1 = n, (a0 ' + x0[m0, a])[n] = (a0 ' + x0)[n], x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)