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Proof step 5 from proof put-del1-in (previous step)

m ≠ 0 ∧ a0 ' + x0 ≠ [], m ≤ # x0, ∀ m, a. x0[m] ≠ a ∧ m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ⊦ a ∈ ((a0 ' + x0).first ' + (a0 ' + x0).rest[m -1, a]) -1l a ↔ a = a0 ∨ a ∈ x0, (a0 ' + x0)[m] = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m ≤ # x0, ∀ m, a. x0[m] ≠ a ∧ m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ⊦ (a0 ' + x0)[m] = a, m = 0, a ∈ (a0 ' + x0[m - 1, a]) -1l a ↔ a = a0 ∨ a ∈ x0

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%pred-03: ⊦ n -1 = n - 1; from specification nat%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-08: ⊦ (a ' + x).first = a; from specification list