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Proof step 5 from proof put-at (previous step)

¬(m ≠ 0 ∧ a0 ' + x0 ≠ []), m ≤ # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m = n, (a0 ' + x0)[m, a][n] = (a0 ' + x0)[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ 0 = n, (a ' + x0)[n] = (a0 ' + x0)[n]

Rule argument

Additional information

Used simplifier rules: put-zero: ⊦ (b ' + x)[0, a] = a + x;%lez: ⊦ 0 ≤ m; from specification nat%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list