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Proof step 4 from proof rtc-one (previous step)

p(a, b) ⊦ ∃ n. rtc(n, a, b)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: rtc1: ⊦ rtc(1, a, b) ↔ p(a, b); Used terms: 1