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Proof step 16 from proof rtc-last (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 17:
Pelem(c, b), Pelem(a, c0), rtc(m0, Pelem)(c0, c), rtc(m0, Pelem)(c, b) ∧ Pelem(b, c0) → (∃ c1. Pelem(c, c1) ∧ rtc(m0, Pelem)(c1, c0)), ∀ Pelem, a, c, b. rtc(m0, Pelem)(a, c) ∧ Pelem(c, b) → (∃ c. Pelem(a, c) ∧ rtc(m0, Pelem)(c, b)) ⊦ ∃ c1. Pelem(a, c1) ∧ (∃ c. Pelem(c1, c) ∧ rtc(m0, Pelem)(c, b))

Rule argument

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

Additional information

Used terms: Pelem, c, b, c0