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Proof step 5 from proof to-unwind-end-fin (previous step)

m0 +1 ≤ # x0, ∀ m. m < # x0 → x0 to m + 1 = (x0 to m) + x0[m] ' ∨ x0 = [], ∀ x0, a. m0 ≤ # x0 ∧ (∀ m. m < # x0 → x0 to m + 1 = (x0 to m) + x0[m] ' ∨ x0 = []) → a ' + (x0 to m0) = (a ' + x0 to m0) + (a ' + x0)[m0] ' ⊦ a ' + (x0 to m0 +1) = (a ' + x0 to m0 +1) + (a ' + x0)[m0 +1] '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m0 < # x0, ∀ x0, a. m0 ≤ # x0 ∧ (∀ m. m < # x0 → x0 to m + 1 = (x0 to m) + x0[m] ' ∨ x0 = []) → a ' + (x0 to m0) = (a ' + x0 to m0) + (a ' + x0)[m0] ', ∀ m. m < # x0 → x0 to m + 1 = (x0 to m) + x0[m] ' ⊦ x0 to m0 + 1 = (x0 to m0) + x0[m0] ', x0 = []

Rule argument

Additional information

Used simplifier rules: to-succ-cons-l-lem: ⊦ a ' + x to n + 1 = a + (x to n);% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ (a ' + x)[n + 1] = x[n]; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)% ⊦ m +1 = m + 1; (s)