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Proof step 3 from proof max-ge-every (previous step)

∀ a. ¬(a ∈ x0 ∧ max(x0) < a), a ∈ a0 + x0, max(a0 + x0) < a ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
max(a0 ' + x0) < a, a ∈ x0, ∀ a1. ¬ a1 ∈ x0 ∨ ¬ max(x0) < a1 ⊦ a = a0

Rule argument

Additional information

Used simplifier rules: max-ge-first: ⊦ ¬ ¬ max(a ' + x) < a ↔ false;%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list