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

(a2 ' + y) -1l a = a3 ' + y0, ordered≤(a3 ' + y0), ordered≤(a2 ' + y), a3 < a0, ∀ a1. ordered≤((a2 ' + y) -1l a1) ⊦ a = a0, a2 < a0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del inst: name: del1-n seq: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a) subst: [a, x, b] → [a, y, a2] flags: rewrite

Additional information

Used simplifier rules: le-r: ⊦ ordered≤(a ' + b ' + x) ↔ ¬ b < a ∧ ordered≤(b + x);%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%del1-y: ⊦ (a ' + x) -1l a = x; from specification list-del%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list Used terms: a, y, a2