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Proof step 19 from proof del1-01-02 (previous step)

x0 = a2 + y, a < a0, ordered≤(x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = a0, ins≤(a, x0) = a ' + x0, x0 = [], a2 < a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
a < a0, ordered≤(a2 ' + y), ∀ a1, b. a1 = b ∨ ins≤(a1, (a2 ' + y) -1l b) = ins≤(a1, a2 ' + y) -1l b ⊦ a = a0, ins≤(a, a2 ' + y) = a ' + a2 ' + y, a2 < a0

Rule argument

Additional information

Used simplifier rules: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x);%app-01: ⊦ ordered≤(x + y) → ordered≤(y);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)