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Proof step 14 from proof put-in-01 (previous step)

m = m0 + 1, m ≤ # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0 -1l x0[m]) ⊦ m = 0, (a0 ' + x0)[m] = a0, a ∈ x0[m0, b] ↔ a = b ∨ a ∈ x0 -1l (a0 ' + x0)[m], a = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
m0 < # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0 -1l x0[m]) ⊦ a ∈ x0[m0, b] ↔ a = b ∨ a ∈ x0 -1l x0[m0], x0[m0] = a0, a = a0, x0 = []

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%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%a: ⊦ (x + y) + z = x + y + z; from specification list%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