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Proof step 8 from proof rtc-last (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 9:
p(a, c), p(c, c0), rtc(m0, c0, b), p(c, c0) ∧ rtc(m0, c0, b) → (∃ c0. rtc(m0, c, c0) ∧ p(c0, b)) ⊦ ∃ c1. (∃ c. p(a, c) ∧ rtc(m0, c, c1)) ∧ p(c1, b)

Rule argument

pos: left 4 terms: c, c0, b abort: F discard: T computed: T

Additional information

Used terms: c, c0, b