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Proof step 3 from proof ins-less-first (previous step)

x = a0 + y, ordered≤(x) ⊦ x = [], ins≤(a, x) = a0 ' + ins≤(a, y), a < a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ordered≤(a0 ' + y) ⊦ ins≤(a, a0 ' + y) = a0 ' + ins≤(a, y), a < 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%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list