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Proof step 7 from proof put-in-02 (previous step)

m = m0 + 1, m ≤ # x0, ∀ m, a. m < # x0 → a ∈ x0[m, a] ⊦ m = 0, a = a0, a ∈ x0[m0, a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
m0 < # x0, ∀ m, a. m < # x0 → a ∈ x0[m, a] ⊦ a = a0, x0 = [], a ∈ x0[m0, a]

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2