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Proof step 2 from proof r1-refl (previous step)

I(pm, s), ∀ s, s0. (Pre1(pm, s) ∧ R1(pm, s, s0) → Pre1(pm, s0)) ∧ (R1(pm, s, s0) ∧ I(pm, s) → I(pm, s0)) ∧ (R2(pm, s, s0) ∧ I(pm, s) → I(pm, s0)) ∧ (Pre2(pm, s) ∧ R2(pm, s, s0) → Pre2(pm, s0)) ∧ (G1(pm, s, s0) ∧ I(pm, s) → R2(pm, s, s0)) ∧ (G2(pm, s, s0) ∧ I(pm, s) → R1(pm, s, s0)), ∀ s, s0, s1. (Pre1(pm, s) ∧ Post1(pm, s, s0) ∧ I(pm, s0) ∧ R1(pm, s0, s1) → Post1(pm, s, s1)) ∧ (Pre2(pm, s) ∧ Post2(pm, s, s0) ∧ I(pm, s0) ∧ R2(pm, s0, s1) → Post2(pm, s, s1)) ∧ (R1(pm, s, s0) ∧ R1(pm, s0, s1) ∧ I(pm, s) → R1(pm, s, s1)) ∧ (R2(pm, s, s0) ∧ R2(pm, s0, s1) ∧ I(pm, s) → R2(pm, s, s1)), ∀ s. (I(pm, s) → R1(pm, s, s) ∧ R2(pm, s, s)) ∧ (Pre1(pm, s) → I(pm, s)) ∧ (Pre2(pm, s) → I(pm, s)) ⊦ R1(pm, s, s)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: s

Additional information

Used terms: s