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Proof step 8 from proof rtc+-unfold-last (
previous step
)
p(a, c), rtc(n0, c, b), rtc(n0, c, b) → n0 = 0 ∨ (∃ c0. p*(c, c0) ∧ p(c0, b)) ⊦ ∃ c. p*(a, c) ∧ p(c, b)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 9
:
p(a, c), rtc(n0, c, b), n0 = 0 ∨ p*(c, c0) ∧ p(c0, b) ⊦ ∃ c. p*(a, c) ∧ p(c, b)
Rule argument
Additional information