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Proof step 10 from proof del1-01-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
ordered≤(a0 ' + x0), ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = a ' + ((a0 ' + x0) -1l b), a0 < a

Rule argument

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;%del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); from specification list-del%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list