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Proof step 5 from proof put-in-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m ≤ # x0, ∀ m, a. m < # x0 → a ∈ x0[m, a] ⊦ m = 0, a = a0, a ∈ x0[m - 1, a]

Rule argument

Additional information

Used simplifier rules: in-06: ⊦ ¬ 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%c-08: ⊦ (a ' + x).first = a; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list