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Proof step 26 from proof rmpos-get (previous step)

x0 = a + y, m1 < # x0, m1 + 1 < # x0, m + 1 < # x0, m1 ≤ m, ∀ n0, m0. n0 + 1 < # x0 ∧ m0 < # x0 → (x0 -p m0)[n0] = (n0 < m0 ⊃ x0[n0];y[n0]) ⊦ (x0 -p m1)[m] = y[m], x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
m1 < # y, m < # y, m1 ≤ m, ∀ n, m0. n < # y ∧ ¬ # y < m0 → ((a ' + y) -p m0)[n] = (n < m0 ⊃ (a ' + y)[n];y[n]) ⊦ ((a ' + y) -p m1)[m] = y[m], y = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)