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

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

User applied rule all left

And obtained the following new goals

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

Rule argument

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

Additional information

Used terms: Pelem, c, c0, b