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Proof step 27 from proof rtc+-unfold-last (previous step)

p(c, b), p(a, c0), rtc(n0, c0, c), ∀ c, b, a. p(c, b) ∧ rtc(n0, a, c) → (∃ c. p(a, c) ∧ rtc(n0, c, b)) ⊦ ∃ c1. p(a, c1) ∧ (∃ c. p(c1, c) ∧ rtc(n0, c, b))

User applied rule all left

And obtained the following new goals

Proof step 28:
p(a, c0), p(c, b), rtc(n0, c0, c), p(c, b) ∧ rtc(n0, c0, c) → (∃ c. p(c0, c) ∧ rtc(n0, c, b)), ∀ c, b, a. p(c, b) ∧ rtc(n0, a, c) → (∃ c. p(a, c) ∧ rtc(n0, c, b)) ⊦ ∃ c1. p(a, c1) ∧ (∃ c. p(c1, c) ∧ rtc(n0, c, b))

Rule argument

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

Additional information

Used terms: c, b, c0