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Proof step 17 from proof propagate (previous step)

H0 ∪ H1 = H3 ∪ H2, !(id, O2, P1)(H3), !(me, O2, P)(H0), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), H3 ⊥ H2, H0 ⊥ H, H1 ⊥ H0 ⊦ id = me, ((!(id, O2, P1) * true))(H1)

User applied rule cut formula

And obtained the following new goals

Proof step 18:
H0 ∪ H1 = H3 ∪ H2, !(me, O2, P)(H0), !(id, O2, P1)(H3), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), H3 ⊆ H1, H1 ⊥ H0, H0 ⊥ H, H3 ⊥ H2 ⊦ id = me, ((!(id, O2, P1) * true))(H1)
Proof step 23:
H0 ∪ H1 = H3 ∪ H2, !(me, O2, P)(H0), !(id, O2, P1)(H3), P0(H), rely(me, O1, H0 ∪ H, O2, H3 ∪ H2), H1 ⊥ H0, H0 ⊥ H, H3 ⊥ H2 ⊦ id = me, ((!(id, O2, P1) * true))(H1), H3 ⊆ H1

Rule argument

fma: H3 ⊆ H1

Additional information

Used formulas: H3 ⊆ H1