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Proof step 12 from proof And-discard (previous step)

true, ∀ s0, P. # s0 < # s → (true → # s0 = 0 ∨ s0 = ∅ ∨ (λ a. P) ∧_ s0 = P) ⊦ # s = 0 ∨ s = ∅ ∨ (λ a. P) ∧_ s = P

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ s0, P. # s0 < # s → (true → # s0 = 0 ∨ s0 = ∅ ∨ (λ a. P) ∧_ s0 = P) ⊦ (λ a. P) ∧_ s = P, s = ∅, # s = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]