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Proof step 10 from proof rtc-last (previous step)

p(a, c), p(c, c0), rtc(m0, c0, b), rtc(m0, c, c1), p(c1, b) ⊦ ∃ c1. (∃ c. p(a, c) ∧ rtc(m0, c, c1)) ∧ p(c1, b)

User applied rule exists right

And obtained the following new goals

Proof step 11:
p(a, c), p(c, c0), rtc(m0, c0, b), rtc(m0, c, c1), p(c1, b) ⊦ (∃ c. p(a, c) ∧ rtc(m0, c, c1)) ∧ p(c1, b)

Rule argument

pos: right 1 terms: c1 abort: F discard: T computed: T

Additional information

Used terms: c1