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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ 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