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Proof step 6 from proof del1-01-02 (previous step)

a0 < a, 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) = (a0 ' + ins≤(a, x0)) -1l b

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a)
Proof step 8:
b ≠ a0, a0 < a, 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)) = (a0 ' + ins≤(a, x0)) -1l b

Rule argument

spec: list-del inst: name: del1-n seq: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a) subst: [a, x, b] → [b, x0, a0] flags: rewrite

Additional information

Used simplifier rules: eq: ⊦ ins≤(a, x) = ins≤(a, y) ↔ x = y;%del1-y: ⊦ (a ' + x) -1l a = x; from specification list-del%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: b, x0, a0