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Proof step 12 from proof ordered-append (previous step)

ordered≤(a ' + x1), ordered≤(x1 + a0 ' + y), ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.first < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], #(a ' + x1 + a0 ' + y) ≤ 1 ∨ ¬ (a ' + x1 + a0 ' + y).rest.first < (a ' + x1 + a0 ' + y).first ∧ ordered≤((a ' + x1 + a0 ' + y).rest), a0 < x1.last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
ordered≤(a ' + x1), ordered≤(x1 + a0 ' + y), x1.first < a, ∀ x. ordered≤(x) ∧ (x ≠ [] → ¬ x.first < x1.last) → ordered≤(x1 + x) ⊦ x1 = [], a0 < x1.last

Rule argument

Additional information

Used simplifier rules: c-08: ⊦ (a ' + x).first = a; from specification list%first-01: ⊦ x ≠ [] → (x + y).first = x.first; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%le-num-06: n0 < n ⊦ ¬ m + (* n) ≤ (* n0); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len: ⊦ #(x + y) = # x + # y; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list