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Proof step 4 from proof max-ge-every (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
max(a0 ' + x0) < a, a ∈ x0, ∀ a1. ¬ max(x0) < a1 ∨ ¬ a1 ∈ x0 ⊦ 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% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)