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Proof step 6 from proof RG-restr-sym (previous step)

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. (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)), Pre1(pm, s1) ∧ Post1(pm, s1, s2) ∧ I(pm, s2) ∧ R1(pm, s2, s3) → Post1(pm, s1, s3), Pre2(pm, s1) ∧ Post2(pm, s1, s2) ∧ I(pm, s2) ∧ R2(pm, s2, s3) → Post2(pm, s1, s3), R1(pm, s1, s2) ∧ R1(pm, s2, s3) ∧ I(pm, s1) → R1(pm, s1, s3), R2(pm, s1, s2) ∧ R2(pm, s2, s3) ∧ I(pm, s1) → R2(pm, s1, s3), ∀ 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)) ⊦ (I(pm, s4) → R2(pm, s4, s4) ∧ R1(pm, s4, s4)) ∧ (Pre2(pm, s4) → I(pm, s4)) ∧ (Pre1(pm, s4) → I(pm, s4))

Proof step 6 from proof RG-restr-sym (previous step)

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

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