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Proof step 5 from proof mk-ext (
previous step
)
∀ rs, f. P0(mk(rs, f)) ↔ P1(mk(rs, f)), ∃ rs, f. H = mk(rs, f) ⊦ P0(H) ↔ P1(H)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
∀ rs, f. P0(mk(rs, f)) ↔ P1(mk(rs, f)) ⊦ P0(mk(rs, f)) ↔ P1(mk(rs, f))
Rule argument
Additional information