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Proof step 15 from proof rtc-trans (previous step)

p(a, c0), p*(b, c), rtc(n0, c0, b), rtc(n, c0, c) ⊦ ∃ n. rtc(n, a, c)

User applied rule exists right

And obtained the following new goals

Proof step 16:
p(a, c0), p*(b, c), rtc(n, c0, c), rtc(n0, c0, b) ⊦ rtc(n + 1, a, c), ∃ n. rtc(n, a, c)

Rule argument

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

Additional information

Used terms: n + 1