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Proof step 16 from proof del1-01-02 (previous step)

a0 ' + x0 ≠ [] ∧ (a0 ' + x0).rest ≠ [], ¬ (a0 ' + x0).rest.first < (a0 ' + x0).first ∧ ordered≤((a0 ' + x0).rest), a < a0, ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ a = a0, ins≤(a, x0) = a ' + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
a < a0, ordered≤(x0), ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ a = a0, ins≤(a, x0) = a ' + x0, x0 = [], x0.first < a0

Rule argument

Additional information

Used simplifier rules: c-08: ⊦ (a ' + x).first = a; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list