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

Pelem(a, c), Pelem(c, c0), rtc(m0, Pelem)(c0, b), rtc(m0, Pelem)(c, c1), Pelem(c1, 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 exists right

And obtained the following new goals

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

Rule argument

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

Additional information

Used terms: c1