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Proof step 6 from proof max-ge-every (
previous step
)
x0 ≠ [], (max(x0) < a0 ⊃ a0;max(x0)) < a, a ∈ x0, ∀ a1. ¬ a1 ∈ x0 ∨ ¬ max(x0) < a1 ⊦ a = a0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 7
:
(max(x0) < a0 ⊃ a0;max(x0)) < a, a ∈ x0, ∀ a1. ¬ a1 ∈ x0 ∨ ¬ max(x0) < a1 ⊦ a = a0, x0 = []
Rule argument
Additional information