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Proof step 18 from proof put-at (previous step)

n = m1 + 1, m0 < # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ m0 + 1 = n, x0[m0, a][m1] = (a0 ' + x0)[n], n = 0, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
m0 < # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ x0[m0, a][m1] = x0[m1], m0 = m1, x0 = []

Rule argument

Additional information

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